Question

The average national SAT score is 1119. If we assume a bell-shaped distribution and a standard deviation equal to 206, what percentage of scores will fall between 501 and 1737? Between 707 and 1531? Between 931 and 1325?

Answer 1

Solution: We are given:

$\mu=1119, \sigma =206$

a. what percentage of scores will fall between 501 and 1737?

In order to find the percentage of scores that fall between 501 and 1737, we use the z score formula first:

When x = 501, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{501-1119}{206}=-3$

When x = 1737, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{1737-1119}{206}=3$

Therefore, we have to find $P(-3\leq z \leq 3)$.

From the empirical rule of normal distribution 99.7% of data falls within 3 standard deviation's from mean.

Therefore, 99.7% of scores will fall between 501 and 1737.

b. what percentage of scores will fall between 707 and 1531?

In order to find the percentage of scores that fall between 707 and 1531, we use the z score formula first:

When x = 707, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{707-1119}{206}=-2$

When x = 1531, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{1531-1119}{206}=2$

Therefore, we have to find $P(-2\leq z \leq 2)$.

From the empirical rule of normal distribution 95% of data falls within 2 standard deviation's from mean.

Therefore, 95% of scores will fall between 707 and 1531.

c. what percentage of scores will fall between 931 and 1325?

In order to find the percentage of scores that fall between 931 and 1325, we use the z score formula first:

When x = 931, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{931-1119}{206}=-1$

When x = 1325, we have:

$z=\frac{x-\mu}{\sigma}$

        $=\frac{1325-1119}{206}=1$

Therefore, we have to find $P(-1\leq z \leq 1)$.

From the empirical rule of normal distribution 68% of data falls within 1 standard deviation's from mean.

Therefore, 95% of scores will fall between 707 and 1531.